Question

Determine, nos dois triângulos a seguir, os valores dos ângulos e .

Answer 1

Vamos usar trigonometria no triângulo retângulo :

$\displaystyle \text{Seno} = \frac{\text{cateto oposto } }{\text{hipotenusa}}$

$\displaystyle \text{Cosseno} = \frac{\text{cateto adjacente } }{\text{hipotenusa}}$

$\displaystyle \text{Tangente } = \frac{\text{Seno }}{\text{Cosseno}} \to \text{Tangente } = \frac{\text{cateto oposto }}{\text{cateto adjacente }}$

ângulos notáveis :

$\displaystyle \boxed{\text{Sen}(30^{\circ}) = \frac{1}{2}} \ \ ; \boxed{\text{Sen}(45^{\circ}) = \frac{\sqrt{2}}{2}} \ \ ; \boxed{\text{Sen}(60^{\circ}) = \frac{\sqrt{3}}{2} }$

$\displaystyle \boxed{\text{Cos}(30^{\circ}) = \frac{\sqrt{3}}{2}} \ \ ; \boxed{\text{Cos}(45^{\circ}) = \frac{\sqrt{2}}{2}} \ \ ; \boxed{\text{Cos}(60^{\circ}) = \frac{1}{2} }$

$\displaystyle \boxed{\text{Tg}(30^{\circ}) = \frac{\sqrt{3}}{3}} \ \ ; \boxed{\text{Tg}(45^{\circ}) = 1} \ \ ; \boxed{\text{Tg}(60^{\circ}) = \sqrt{3}}$

Item a)

Cateto adjacente = 18

Hipotenusa = $12\sqrt{3}$

Aplicando Cossseno :

$\displaystyle \text{Cos}(\alpha) = \frac{18}{12\sqrt{3}} \to \text{Cos}(\alpha) = \frac{6}{2\sqrt{3}}$

Racionalizando :

$\displaystyle \text{Cos}(\alpha) = \frac{3}{2\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}} \to \text{Coss}(\alpha) = \frac{3\sqrt{3}}{2.3} \to \text{Cos}(\alpha) = \frac{\sqrt{3}}{2}$

$\huge\boxed{\alpha = 30^{\circ}} \checkmark$

item b)

Cateto oposto = 5

Cateto adjacente = $5\sqrt{3}$

Aplicando Tangente :

$\displaystyle \text{Tg}(\beta) = \frac{5}{5.\sqrt{3}} \to \text{Tg}(\beta) = \frac{1}{\sqrt{3}}$

Racionalizando :

$\displaystyle \text{Tg}(\beta) = \frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}} \to \text{Tg}(\beta) = \frac{\sqrt{3}}{3}$

$\huge\boxed{\beta = 30^{\circ}}\checkmark$